Calculating load requirements

[SavannahLion]

How does one calculate the load requirements of an unknown circuit? I know I'm drawing 5 volts but how does one figure the watts or amperage of an unknown circuit?

[MonMotha]

Power is related to voltage and current: P = V * I.  You need to know 2 of the 3 to compute the other.  If you have a resistive load and you know the resistance R, you additionally know (via Ohm's law) that V = I * R and various substitutions let you come up with P = V^2 / R or P = I^2 * R, but then you need to know R.

Note that many loads are NOT resistive.  many loads are "constant power" - their current increases as the voltage decreases so that P stays roughly constant.  Some loads behave somewhere in between, and some esoteric loads behave really weirdly (like fluorescent tubes).

If you just have a "black box" circuit, you have no idea how voltage will relate to current.  It's possible to build a circuit that does just about anything.

If you have a multimeter, you can measure the current directly.  You have to put the meter "in-line" with the circuit.  That is, you unhook one supply lead, and hook the meter up between them on the current (amperage) setting.  For AC loads, there also exist "clip-on" type meters, though their accuracy varies, and this doesn't work on DC, which it sounds like you've got.

[SavannahLion]

Power is related to voltage and current: P = V * I.  You need to know 2 of the 3 to compute the other.  If you have a resistive load and you know the resistance R, you additionally know (via Ohm's law) that V = I * R and various substitutions let you come up with P = V^2 / R or P = I^2 * R, but then you need to know R.

Yep, that's my problem alright. Not enough info to compute.

Note that many loads are NOT resistive.  many loads are "constant power" - their current increases as the voltage decreases so that P stays roughly constant.  Some loads behave somewhere in between, and some esoteric loads behave really weirdly (like fluorescent tubes).

If you just have a "black box" circuit, you have no idea how voltage will relate to current.  It's possible to build a circuit that does just about anything.

It's not technically black box I guess. It's a circuit I built (will build. If my fiance, kids and animals leave me alone long enough, I might actually finish assembling the breadboard).

If you have a multimeter, you can measure the current directly.  You have to put the meter "in-line" with the circuit.  That is, you unhook one supply lead, and hook the meter up between them on the current (amperage) setting.  For AC loads, there also exist "clip-on" type meters, though their accuracy varies, and this doesn't work on DC, which it sounds like you've got.

As luck would have it, Equus actually has the manual for my DMM online so I was able to look up which of the appropriate switches I should use. Naturally, it's a function I never used until now. Go figure.

I haven't read the instructions entirely yet so this may sound rather obvious.

When I usually build bread board circuits, I usually use a battery to test and power the circuit. In this case, the final circuit is going to be inserted into a PC where the 5v line is specified to be a max 10mA (though one source cites as high as 500mA). Is it acceptable to get a reading using the battery or do I want to drop the circuit into it's final environment then get a reading? I would imagine the reading is going to be the same regardless of the source or am I missing something fundamental here?


fixed bad markup

[danny_galaga]

Keep adding to it until something goes pop. Then you know you should have had one less than that 

[SavannahLion]

Keep adding to it until something goes pop. Then you know you should have had one less than that 

 

[danny_galaga]

Keep adding to it until something goes pop. Then you know you should have had one less than that 

 

I stand by my assertion

[MonMotha]

10mA seems like an AWFULLY low rating for a 5V line on a PC supply.  That's only 50mW (0.05W).  Most PC supplies can handle at least half an amp (2.5W) on their 5V standby (always on) output and MUCH higher (often north of 20A) on their main 5V outputs.

You don't need to measure it in the final installation as long as your supply conditions are roughly the same (i.e. 5V and not bouncing off any current limiters).  If you post the circuit, I can help you analyze it to figure out what it "should" be.

[SavannahLion]

10mA seems like an AWFULLY low rating for a 5V line on a PC supply.  That's only 50mW (0.05W).  Most PC supplies can handle at least half an amp (2.5W) on their 5V standby (always on) output and MUCH higher (often north of 20A) on their main 5V outputs.

Yeah, I want to draw off of the PC standby line.

You don't need to measure it in the final installation as long as your supply conditions are roughly the same (i.e. 5V and not bouncing off any current limiters).  If you post the circuit, I can help you analyze it to figure out what it "should" be.

Err...I'm a little embarrassed to put up an untested circuit. Right now, the only "schematic" I have is the TINA simulation file. Some of the graphics and IC pinouts are a little weird. I can take a screenshot of the circuit from TINA though.

[MonMotha]

Looks like I don't have TINA on here (amazing...), so a screenshot will work fine.

As for the ratings of the 5VSB line, check your PSU.  Figure your motherboard draws ~200-300mA or more if you have USB charging during sleep enabled.

[SavannahLion]

Yeah, my version of TINA is pretty old 6.0 I think. If this laptop craps out, I'm not even sure if I can reinstall. Hell, I don't know if I ever can reinstall. I have no idea where the installation files are.

[MonMotha]

TI has a free version of TINA (TINA-TI) on their website.  Probably sufficient for your use.  There's also LTSpice, which is what I usually use.

[SavannahLion]

That's weird.... what happened to the last post?

[SavannahLion]

Well  

Dunno what happened and I'm in no mood to recreate the post so here's the picture again.

I think I have an improved version in the works.

I tried to finish up this circuit but I discovered that I lack the key RC combo to make the whole damn thing work. I also ran out of 10k resistors. Checked Metro Electronics and they want $7 for 200 resistors. ugh.....

[MonMotha]

Sorry I never got around to looking at this.  I haven't had the time to actually determine if your circuit works, but at a quick glance, I wouldn't worry too much about the power consumption.  Probably no more than a few dozen mA.

[SavannahLion]

No biggie. While at work, I get so bored that I created a new version of this circuit and created two others that don't use a 555. They seem to work OK in the simulator (though if I don't use a 555. there's always a delay on switch open I can't quite resolve) Now if I can find the time to actually put a RW circuit together and see just how accurate my simulations are, I'll be gold.

[MonMotha]

I'd just use a small micro and an open-drain output.